The electric field on the axis of a uniformly charged ring has magnitude 400kN/C

Question

The electric field on the axis of a uniformly charged ring has magnitude 400kN/C at a point 5.8cm from the ring center. The magnitude 16 cm away from the center is 110kN/C; in both cases the field points away from the ring. Determine: a) The radius of the ring. b) Its charge. Please solve with work using THESE numbers.

Explanation / Answer

The formula for the electric field intensity at a distance x from the center of the ring of radius a and having charge Q on its axis is E = kxQ/(x^2 + a^2)^(3/2) => 365000 = 9 x 10^9 * (0.058) * Q / [(0.058)^2 + a^2] and 110000 = 9 x 10^9 * (0.23) * Q / ([(0.23)^2 + a^2] Taking ratio, 365/110 = (5.8)/(23) * [(0.23)^2 + a^2]/[(0.058)^2 + a^2] => 13.1583 = [(0.23)^2 + a^2]/[(0.058)^2 + a^2] => (13.1583) * (a^2 + 0.003364) = (a^2 + 0.0529) => 12.1583 a^2 = 0.008635 => a = 0.02665 m => a = 2.665 cm. Charge Q = 110000 * {0.0529 + (0.02665)^2] / (9 x 10^9 x 0.23) C => Q = 110000 * (0.05361) / (2.07 x 10^9) C => Q = 2.849 x 10^-6 C = 2.849 micro-C

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